HDU 1573 x problem (unary linear congruence equation set)

X problem Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 3850 Accepted Submission (s): 1228


Problem description The number of x in a positive integer less than or equal to n is satisfied: x mod a[0] = b[0], x mod a[1] = b[1], x mod a[2] = b[2], ..., x mod a[i] = b[i], ... (0 < A[i] <= 10).
The first behavior of input data is a positive integer t, which indicates that there is a T group of test data. The first behavior of each set of test data is two positive integers n,m (0 < N <= 1000,000,000, 0 < M <= 10), indicating that x is less than or equal to N, and that each has M elements in arrays A and B. The next two lines, each with m positive integers, are the elements in A and B, respectively.
Output corresponds to each set of inputs, outputting a positive integer in a separate row that represents the number of x that satisfies the condition.
Sample Input

310 31 2 30 1 2100 73 4 5 6 7 8 91 2 3 4 5 6 710000 101 2 3 4 5 6 7 8 9 100 1 2 3 4 5 6 7 8 9

Sample Output

103

Idea: or the M-congruence equation 22 combined with EXTGCD in turn to solve

249MS 1400K 1068 B g++ #include <cstdio> #include <iostream> #include <cstring> #include <algorithm >using namespace Std;typedef Long long ll;ll a[15],b[15];ll n,m;ll x,y;ll e_gcd (ll a,ll b,ll&x,ll&y) {ll ans    ;    if (b==0) ans=a,x=1,y=0;    else ANS=E_GCD (b,a%b,y,x), y-= (A/b) *x; return ans;}    BOOL Flag;int Main () {int T;    scanf ("%d", &t);        while (t--) {flag=0;        scanf ("%i64d%i64d", &n,&m);        for (int i=1;i<=m;i++) scanf ("%i64d", &a[i]);        for (int i=1;i<=m;i++) scanf ("%i64d", &b[i]);            for (int i=2;i<=m;i++) {ll a=a[i-1],b=a[i],r=b[i]-b[i-1];            ll D=E_GCD (a,b,x,y);            if (r%d) flag=1;            ll t=b/d;            x= (x*r/d%t+t)%t;        a[i]=a[i-1]*a[i]/d;//merged into a new congruence equation b[i]=x*a[i-1]+b[i-1];//merged into a new congruence equation} if (flag) puts ("0");            else {ll ans=0,lcm=a[m]; b[m]= (b[m]%lcm+lcm-1)%lcm+1;//the solution in 1~lcm[a1,a2,a3...an] Within, since the title requirement is positive not including 0 while (b[m]<=n) ANS++,B[M]+=LCM;        printf ("%i64d\n", ans); }} return 0;}

HDU 1573 x problem (unary linear congruence equation set)